fastJson 将对象转换为 json 字符串

V2EX = way to explore

V2EX 是一个关于分享和探索的地方

已注册用户请登录

这是一个创建于 3157 天前的主题，其中的信息可能已经有所发展或是发生改变。

测试代码：

// jackson
private static ObjectMapper om = new ObjectMapper ();
public static String objectToString (Object obj ) throws IOException{
    return om.writeValueAsString (obj );
}

@Test
public void test1 () throws IOException {
    Map<String, Object> map = new HashMap<String, Object>();

    Map<String, String> map2 = new HashMap<String, String>();
    map2.put ("name", "1");
    map2.put ("age", "12");
    map.put ("att", map2 );

    List<String> list = new ArrayList<String>();
    list.add ("hi");list.add ("hello");
    map.put ("list", list );

    List<Map<String, String>> confuselist = new ArrayList<Map<String, String>>();
    Map<String, String> map3 = new HashMap<String, String>();
    map3.put ("name", "1");
    map3.put ("age", "12");
    confuselist.add (map3 );
    confuselist.add (map2 );
    map.put ("confuselist", confuselist );

    Object array = Array.newInstance (String.class, 10 );
    Array.set (array, 0, "a");
    Array.set (array, 1, "b");
    Array.set (array, 2, "c");
    map.put ("array", array );
    // 使用 jackson
    System.out.println (objectToString (map ));
    // 使用 fastJson
    System.out.println (JSON.toJSONString (map ));
}

输出：

{"att":{"age":"12","name":"1"},"list":["hi","hello"],"confuselist":[{"age":"12","name":"1"},    {"age":"12","name":"1"}],"array":["a","b","c",null,null,null,null,null,null,null]}

{"att":{"age":"12","name":"1"},"list":["hi","hello"],"confuselist":[{"age":"12","name":"1"},{"$ref":"$.att"}],"array":["a","b","c",null,null,null,null,null,null,null]}

fastJson 对相同的对象使用引用标记节省字符串体积吗？有没有办法不用引用。

6 条回复 • 2015-09-02 00:35:52 +08:00